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Department of Civil Engineering-I.I.T. Delhi

CEL 212: Environmental Engineering Second Semester 2011-2012

Home Work 9 (Disinfection) Q1. The Chick’s Law: For disinfection, assume N (0) is initial number of pathogens and N (t) is remaining

number of pathogens at time and given by: N (t) =N (0) × exp (-K×t); where K is disinfection rate

(unit=1/unit of time) and depends on disinfectant-pathogen interaction and solution characteristics. Here R-

log removal: R=-log10 [N (t)/N (0)].

If disinfectant concentration (Cdisinfectant) and contact time (tc) are related to each other by following equation

(the Watson’s Law):

(Cdisinfectant)n × (tc) = constant (standard unit: C in mg/L and tc in minute)

Calculate R-log removal value for 99.9% removal? What is the remaining pathogen concentration at this

removal after 1 minutes of contact time? (Assume K=0.046/min) [3+3=6 points]

Solution: R-log removal = -log10 [1-Nt/N0]

Given removal = 99.9%, so Nt/N0= (1-99.9/100) =0.001

R-log removal = -log10 [0.001] =3 (answer)

For calculating remaining fraction of pathogens after 1 minute of disinfection with K=0.046/min, use the Chick’s Law: Nt/N0 = exp (-k ×t) = exp (-0.046/min×1 min) =0.9550 (i.e., 95.50%) (answer)

Q2. An experiment shows that a concentration of 0.1g/m3 of free available chlorine yield a 99% kill of

bacteria in 8 minutes. What contact time is required to achieve a 99.9% kill at a free available chlorine

concentration of 0.05 g/m3? Assume that Chick’s Law and Watson’s Law hold with n=1. [2+2=4 points]

Solution: Given: For 99% kill: C= 0.1 g/m3 and time (t) =8 minutes

Chick’s Law: Nt=N0×exp (-k ×t)

Calculation of disinfection rate constant:

Nt/N0= (1-99/100) =0.01 in 8 minutes

From Chick’s Law: 0.01 = exp (-k ×8) => k = - (1/8) ln (0.01) = 0.5756/min (answer)

Using calculated k value, calculate time for getting 99.9% kill:

Nt/N0= (1-99.9/100)=0.001

Using Chick’s Law: 0.001 = exp (-0.5756×t) =>t = - (1/0.5756) ln (0.001) =12 min (answer)

Note: Watson’s Law: Cn×t=constant = > C×t=constant (as n=1)

For 99.9% kill: C= 0.1 g/m3 and time (t) =8 minutes. So, C×t value =(0.1 *1000mg/1000L)*(12 minutes)

= 1.2 (mg/L)(min.)

To determine contact time using 0.05 g/m3, Ct is equal for both cases.

1.2 (mg/L)(min.) = (0.05 ×1000mg/1000L)*(t minutes)

t= 24 min. (answer)

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Q3. For wastewater consists of ammonia, organic matter and microorganisms), draw breakthrough curve

using following information and answer following questions:

Chlorine dosage (mg/L) 0.1 0.5 1.0 1.5 2.0 2.5 3.0

Chlorine residual (mg/L) 0.0 0.4 0.8 0.4 0.4 0.9 1.4

(i) Discuss the significance of different regions. [8 points]

(ii) Calculate chlorine dose to achieve 0.75 mg/L free available chlorine? [4 points]

Solution:

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Q4. Look at the following relationship between concentration of free residual chlorine and contact time

required for 99% kill (Watson’s Law: C0.86

tp= λ (constant) for different pathogens).

Pathogen type Adenovirus 3 E.coli Coxsackievirus A2

λ (constant) 0.098 0.24 6.3

For given chlorine dose, how long would you like to disinfect to achieve maximum removal of all pathogens?

Explain the result. [6+4=10 points]

Solution: For given chlorine dose: high contact time is required for high λ value. So we need longest contact time

for Coxsackievirus A2 than that for other pathogens. Thus to achieve maximum removal of all pathogens we need

contact time equal to that of Coxsackievirus A2.

Q5. Comment on decay of adenovirus (survival = Nt/N0) using low pressure (LP) and medium pressure

(MP) UV rays. Note UV dose is given in milli joules/cm2.

Solution: Example: Log (survival)= (-4) (i.e., 4-log removal)

Log (Nt/N0)=-4; Nt=N0 (10-4) or Removal = (1-0.0001)100= 99.99%

LP UV rays produce high decay of adenovirus compared to MP UV rays. In initial regions, LP UV rays give high

disinfection rate than MP UV rays, but for high dose, both produces similar extent of virus decay.

Q6. Comment on effect of light and dark on concentration of five fecal indicators. Concentration values are

shown in box plots where middle line show median value, below 25th

percentile value (i.e., 25% of values

lower than this) and above 75th

percentile.

Solution: Dark conditions give higher CFU values, i.e., lesser removal of pathogens than that due to light conditions.

Highest removal was observed for Enterococci than other microorganisms as lowest remaining concentration.

Lowest removal was observed for F-RNA phages than other microorganisms.

Higher removal for bacterial indicator (i.e., for fecal coliforms) was observed than viral indicators (i.e., for somatic

coliphage and F-RNA phages).

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Q7. Comment on effect of UV dose on virus survival for two sets of data. Note that two linear models are fit

to data (this is first order data).

Solution: Adeno 2 virus is disinfected at slower rate than other pathogens with increasing UV dose. For high log-removal,

high UV dose is required for adeno 2 virus than other pathogens.

Both linear models appear to fit the observed decay data, however, it is difficult to comment on model fitting due

to difficulty in determining goodness-of-fit of fit models to data.

Q8. Comment on dose requirements for inactivation of viruses at different levels by UV light for each of

viruses studied.

Solution: With high UV dose, high log-removal can be achieved.

For 99.99% removal (i.e., for 4-log removal), highest UV dose is required for adenovirus type 2 than other

pathogens with lowest dose required is for echovirus 2.

These data also indicate that adenovirus type 2 is most resistant to UV dose (i.e., hard to kill). Further, adenovirus

2 requires relatively higher UV dose than other pathogens (at least 4-5 times higher).